算法查找素用最少量的计算 [英] Algorithm for finding a prime with the least amount of computations
问题描述
假设你要编写一个函数/方法找到一个素数,这将是最有效的方式做到这一点?我想,这将是一个测试,是这样的:
Assuming you were going to write a function / method to find a prime number, what would be the most efficient way to do this? I'm thinking that it would be a test that is something like this:
code以下在半C ++
Code Below in semi-c++
bool primeTest (int x) { //X is the number we're testing
int testUpTo = (int)((sqrt(x))+1);
for (int i=3; i<testUpTo; i+=2){
if ((x%i)==0) {
return false;
}
}
return true;
}
是否有人有一个更好的办法去解决这一点,将需要较少的计算?
Does someone have a better way to go about solving this that will take less computations?
编辑:改变code咯,两次。我并没有考虑任何特定的语言写这篇文章,但我想这是C ++对Java由于字布尔。
edit: Changed code slightly, twice. I didn't write this with any specific language in mind, although I suppose it's C++ over java due to the word bool.
推荐答案
我会使用米勒拉宾测试时,其可以容易地作出确定性为编号大于341,550,071,728,321小(2 ^ 31比小得多)。
I would use the Miller Rabin test, which can easily be made deterministic for numbers smaller than 341,550,071,728,321 (and 2^31 is much smaller than that).
伪code:有若干不同的情况
Pseudocode: there are a number of different cases.
-
X
小于9:返回(X放大器; 1)= 0 || x == 2
-
X
超过约200小(可修改):使用审判庭(你使用的是什么) -
X
比1373653更小:用米勒拉宾与基地2和3 -
X
小于4759123141(即一切):。用米勒拉宾与基地2,7和61
x
smaller than 9: Return(x & 1) != 0 || x == 2
x
smaller than about 200 (tweakable): use trial division (what you used)x
smaller than 1373653: use Miller Rabin with bases 2 and 3.x
smaller than 4759123141 (that is everything else): use Miller Rabin with bases 2, 7 and 61.
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